Question

Ice cream is being put into a cone at a rate of 4 cm3/s. The cone has a radius of 2 cm and a height of 8 cm.

Answer 1

Answer:

0.5052 cm per sec ( approx )

Step-by-step explanation:

∵ The volume of a cone is,

$V=\frac{1}{3}\pi (r)^2h$

Where,

r = radius,

h = height,

Here, r = 2 and h = 8,

Thus, the volume of the cone is,

$V=\frac{1}{3}\pi (2)^2 8=\frac{32\pi }{3}$

When the cone is half filled,

Volume would be,

$V_1=\frac{V}{2}=\frac{16\pi }{3}$

Now, $\frac{r}{h}=\frac{1}{4}\implies r=\frac{h}{4}$

$V_1=\frac{1}{3}\pi (\frac{h}{4})^2 h=\frac{\pi h^3}{48}----(1)$

$\implies \frac{\pi h^3}{48} = \frac{16\pi }{3}$

$\implies h \approx 6.35$

Differentiating equation (1) with respect to t ( time )

$\frac{dV_1}{dt}=\frac{\pi h^2}{16}\frac{dh}{dt}$

We have, $\frac{dV_1}{dt}=4\text{ cube cm per s},h=6.35\text{ cm}$

$4=\frac{\pi (6.35)^2}{16}\times \frac{dh}{dt}$

$\implies \frac{dh}{dt}=\frac{64}{6.35^2 \pi}\approx 0.5052\text{ cm per sec}$